need help with math

Question

I've been stuck on these problems for a while.

1. Use the given transformation to evaluate the integral. ∫∫R 8xy dA, where R is the region in the first quadrant bounded by the lines y = 1x/3 and y = 2x and the hyperbolas xy = 1/3 , xy = 2; x = u/v , y = v.

My answer was (154/9)ln(9/2) but it's wrong.

1. Evaluate the given integral by making an appropriate change of variables. ∫∫R 2((x − 5y)/(4x − y)) dA, where R is the parallelogram enclosed by the lines x − 5y = 0, x − 5y = 4, 4x − y = 7, and 4x − y = 8.

My answer for this was (32/19)ln(10/19) which was also incorrect.

Help would be really appreciated.

Answer 1

Problem 1. The given transformation x = u/v, y = v

The inverse transformation u = xy, v = y

The region is bounded by curves xy = 1/3, xy = 2, y = x/3, and y = 2x

In uv-coordinates these curves transforms to u = 1/3, u = 2, v = √(u/3), and v = √(2u).

So, in uv-coordinates the region transforms to { 1/3 ≤ u ≤2, √(u/3) ≤ v ≤ √(2u) }

Function 8xy = 8u

Jacobian:

∂x/∂u ∂x/∂v 1/v -u/v²

= = 1/v

∂y/∂u ∂y/∂v 0 1

So, ∫∫R 8xy dA = _1/3 ∫ ² du _√(u/3) ∫ ^√(2u)√ dv (8u)(1/v) = _1/3 ∫ ² du 8u _√(u/3) ∫ ^√(2u) dv (1/v)

_√(u/3) ∫ ^√(2u) dv (1/v) = ln(√(2u)) - ln(√(u/3)) = ln(√((2u)/(u/3)) = ln(√6) = (ln 6)/2

_1/3 ∫ ² (8u)(ln 6)/2 du = 4 ln 6 _1/3 ∫ ² u du = 4 ln 6 [ 2²/2 - (1/3)²/2] = 2 ln 6 (4 - 1/9) = (70/9) ln 6

Problem 2. Find convenient transformation.

Lines x − 5y = 0, x − 5y = 4, 4x − y = 7, and 4x − y = 8 gives the idea: u = x - 5y and v = 4x - y

Then the parallelogram will transform to the rectangle { 0 ≤ u ≤ 4, 7 ≤ v ≤ 8}

The function 2(x − 5y)/(4x − y) = 2u/v.

To find Jacobian we need to express x and y from u and v.

Solve the linear system u = x - 5y for x and y.

v = 4x - y

I skip the calculations (convenient to use elimination method), but the solution is

x = -u/19 + 5v/19

y = -4u/19 + v/19

Jacobian:

∂x/∂u ∂x/∂v -1/19 5/19

= = -1/19² - (-20/19²) = 19/19² = 1/19

∂y/∂u ∂y/∂v -4/19 1/19

So, ∫∫R 2((x − 5y)/(4x − y)) dA = ₀∫⁴ du ₇∫⁸ dv (2u/v)(1/19) = ₀∫⁴ du (2u/19) ₇∫⁸ (1/v) dv

₇∫⁸ (1/v) dv = ln v |₇ ⁸ = ln 8 - ln 7 = ln(8/7)

₀∫⁴ du (2u/19) ln(8/7) = (2/19) ln(8/7) ₀∫⁴ u du = (2/19) ln(8/7) 8 = (16/19) ln(8/7)